Question

Consider the following solutions: 0.010 m Na3PO4 in water 0.020 m CaBr2 in water 0.020 m KCl in water

Answer 1

Answer:

a) KI and Na₃PO₄

b) HF

c) CaBr₂

Explanation:

The depression in freezing point, relative lowering of vapor pressure and elevation in boiling point are colligative properties.

The collegative properties depend upon the number of solute particles.

Let us see the concentration of particles in each of the given solutions:

a) 0.010 m Na₃PO₄ in water : each molecule will give three sodium ions and one phosphate ion.

So the concentration of particles in the solution = 4 X 0.01 = 0.04 m

b) 0.020 m CaBr₂ in water : each molecule will give one calcium ion and two bromide ion thus the concentration of particles in the solution

= 3 X 0.02 = 0.06 m

c) 0.020 m KCl in water: each molecule gives two ions thus the concentration of solution = 2 X 0.02 = 0.04 m

d) 0.020 m HF (weak acid) in water: the acid is weak, so each molecule will not give two ions completely (complete dissociation will not be observed). concentration of solution will be more than 0.02 but less than 0.04.

a)  Both KCl and Na₃PO₄ will have same boiling point as 0.04 C₆H₁₂O₆.

b) the solution with least concentration of solute will have highest pressure. Thus HF is the answer,

c) the largest depression will be in the solution with highest solute concentration: Answer: CaBr₂.